A200 g oscillator in a vacuum chamber has a frequency of 2.0 hz. when air is admitted, the oscillation decreases to 60% of its initial amplitude in 50 s. how many oscillations will have been completed when the amplitude is 30% of its initial value?

To solve this given problem, we make use of the formula:

A = Ao e^( − b t / 2 m)

Substituting all the given values into the equation:

A / Ao = e^( − b t / 2 m)

When A / Ao = 0.60 and t = 50 s, we find for b:

0.60 = e^( − b t / 2 m)

ln ( 0.60 ) = − b t / 2 m

b = − ( 2 m ) ln ( 0.60 ) / t

b = ( − 2 ) ( .200 ) ln ( 0.60 ) / 50

b = .00409

When A / Ao = 0.30, we find for t:

0.30 = e^( − b t / 2 m)

ln ( 0.30 ) = − (0.00409) t / 2 (0.200)

t = − (0.200) ln ( 0.30 ) / 0.00409

t = 118 s

Therefore the number of oscillations is:

oscillations = f * t = 2 s^-1 (118 s) = 236

236 oscillations

answer; equivalent to a one(1) horsepower motor;

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